Question

Find the derivative of the following functions from first principle:
(i) −x (ii) (-x)-1 (iii) sin (x + 1) (iv) cos (x – $\frac{\pi}{8}$)

Answer 1

(i) Let f(x) = -x. Accordingly, f(x+h)=-(x+h)
By first principle,
f'(x)=$\lim_{h\rightarrow 0}$ $\frac{f(x+h)-f(x)}{h}$
=$\lim_{h\rightarrow 0}$ $\frac{-(x+h)-(-x)}{h}$
=$\lim_{h\rightarrow 0}$ $\frac{-x-h+x}{h}$
=$\lim_{h\rightarrow 0}$ $\frac{-h}{h}$
=$\lim_{h\rightarrow 0}$(-1)=-1

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(ii) Let f(x)=(-x)^-1 = $\frac{1}{-x}$ = $-\frac{1}{x}$ Accordingly, f(x+h)=$-\frac{1}{x+h}$
By first principle,
f'(x)=$\lim_{h\rightarrow 0}$ $\frac{f(x+h)-f(x)}{h}$
=$\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$-\frac{1}{x+h}-(\frac{-1}{x})$]
=$\lim_{h\rightarrow 0}$$\frac{1}{h}$[$\frac{-1}{x+h}+\frac{1}{x}$]
=$\lim_{h\rightarrow 0}$$\frac{1}{h}$[$\frac{-x+(x+h)}{x(x+h)}$]
=$\lim_{h\rightarrow 0}$$\frac{1}{h}$[$\frac{-x+x+h}{x(x+h)}$]
=$\lim_{h\rightarrow 0}$$\frac{1}{h}$[$\frac{h}{x}$(x+h)]
=$\lim_{h\rightarrow 0}$ $\frac{1}{x}$(x+h)
= $\frac{1}{x}$.x - $\frac{1}{x^2}$

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(iii) Let f(x) = sin (x + 1). Accordingly, f(x+h)=sin(x+h +1)
By first principle,
f'(x)=$\lim_{h\rightarrow 0}$ $\frac{f(x+h)-f(x)}{h}$
=$\lim_{h\rightarrow 0}$ $\frac{1}{h}$[sin(x+h+1) -sin(x+1)]
=$\lim_{h\rightarrow 0}$ 1$\frac{1}{h}$[2cos(x+h+1+x+$\frac{1}{2}$) sin(x+h+1-x-1)]
=$\lim_{h\rightarrow 0}$ $\frac{1}{h}$[ 2cos (2x + h+$\frac{2}{2}$) sin($\frac{h}{2}$)]
=$\lim_{h\rightarrow 0}$ $\frac{1}{h}$[ 2cos ($\frac{2x+h+2}{2}$) sin($\frac{h}{2}$)/($\frac{h}{2}$)]
=cos($\frac{2x+0+2}{2}$).1 [$\lim_{h\rightarrow 0}$ $\frac{sin\,x}{x}$ =1]
=cos(x+1)

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(iv) letf(x) cos(x-$\frac{\pi}{8}$). Accordingly, f(x+h)=cos(x+h -$\frac{\pi}{8}$)
By first principle,
f'(x)=$\lim_{h\rightarrow 0}$ $\frac{f(x+h)-f(x)}{h}$
=$\lim_{h\rightarrow 0}$$\frac{1}{h}$[cos(x+h-$\frac{\pi}{8}$)- cos(x-$\frac{\pi}{8}$)]
=$\lim_{h\rightarrow 0}$$\frac{1}{h}$[$\frac{-2sin(x+h-\frac{\pi}{8}+-\frac{\pi}{8})}{2\,sin(\frac{x+h-\frac{\pi}{8}-x+\frac{\pi}{8}}{2})}$]
=$\lim_{h\rightarrow 0}$$\frac{1}{h}$[$\frac{-2sin(2x+h-\frac{\pi}{4})}{2}$ $sin \frac{h}{2}$]
=$\lim_{h\rightarrow 0}$[-sin($\frac{2x+h-\frac{\pi}{4}}{2}$) $\frac{sin\frac{h}{2}}{\frac{h}{2}}$]
=-sin($\frac{2x+0-\frac{\pi}{4}}{2}$) .1
=-sin(x-$\frac{\pi}{8}$)